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I have a solution for the equation $$x^2-3y^2=4\tag{1}$$ i.e. $(2,0)$ which is quite trivial, I also found solution for the equation $$x^2-3y^2=1\tag{2}$$ i.e. $(1,0)$, similarly trivial. I know the identity related with Brahmagupta: $$x_1-Ny_1^2=k_1\tag{3}$$ $$x_2-Ny_2^2=k_2\tag{4}$$ $$k_1k_2=(x_1-N_1y_1^2)(x_2-N_2y_2^2)=(x_1x_2-Ny_1y_2)^2-N(x_1y_2-x_2y_1)^2\tag{5}$$ So given a solution for $(1)$ as $(x_1, y_1)$, and for $(2)$ as $(1,0)$, we do not obtain any new solutions for $(1)$ [$k_1=4,k_2=1$].

Moreover with some such diophantine equations as $x^2-10y^2=9$ we cannot generate all solutions with a single solutions (MathWorld page illustrates how we have 3 seeds for different families of solutions)

So how can I get all positive solutions of $(1)$ and in general how to do so. Finding a solution may be done by the Chakravala method but what later?

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  • $\begingroup$ do you have a sript about the theory which you can use here? $\endgroup$ – Dr. Sonnhard Graubner Jun 9 '17 at 17:23
  • $\begingroup$ @Dr.SonnhardGraubner The identity is from Wikipedia and the Chakravala method was taught to us at school but you can also find it in Wikipedia. Both on this page $\endgroup$ – RE60K Jun 9 '17 at 17:25
  • $\begingroup$ See math.stackexchange.com/questions/648781/…. $\endgroup$ – lhf Jun 9 '17 at 17:27
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The equation $x^2-3y^2=4$ is best studied in $\mathbb Z[\sqrt 3]$ using the norm $N(x+y\sqrt 3)=x^2-3y^2$.

If $N(\alpha)=4$ and $N(\beta)=1$, then $N(\alpha\beta^n)=4$ and so $\alpha\beta^n$ gives you several solutions, as long as $\beta\ne1$. In this case, we can take $\beta = 2+1\sqrt3$. You get the linear recurrence $$ x_{n+1} = 2x_{n} + 3y_n, \qquad x_0 = 2 $$ $$ y_{n+1} = \hphantom{2}x_{n} + 2y_n, \qquad y_0 = 0 $$

You also get solutions if you can find $\gamma$ with $N(\gamma)=-1$. Then $\alpha\gamma^{2n}$ gives you several solutions. But there is no such $\gamma$ because $x^2 \equiv -1 \bmod 3$ has no solutions.

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Beginning with the solutions $(2,0)$ and $(4,2)$ for $x^2 - 3 y^2 = 4,$ all the rest of them come from the linear recurrences $$ x_{n+2} = 4 x_{n+1} - x_n, $$ $$ y_{n+2} = 4 y_{n+1} - y_n. $$

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jagy@phobeusjunior:~$ ./Pell_Target_Fundamental
  Automorphism matrix:  
    2   3
    1   2
  Automorphism backwards:  
    2   -3
    -1   2

  2^2 - 3 1^2 = 1

 u^2 - 3 v^2 = 4

Fri Jun  9 13:36:39 PDT 2017

u:  4  v:  2 ratio: 2  SEED   BACK ONE STEP  2 ,  0
u:  14  v:  8 ratio: 1.75
u:  52  v:  30 ratio: 1.73333
u:  194  v:  112 ratio: 1.73214
u:  724  v:  418 ratio: 1.73206
u:  2702  v:  1560 ratio: 1.73205
u:  10084  v:  5822 ratio: 1.73205
u:  37634  v:  21728 ratio: 1.73205
u:  140452  v:  81090 ratio: 1.73205
u:  524174  v:  302632 ratio: 1.73205
u:  1956244  v:  1129438 ratio: 1.73205
u:  7300802  v:  4215120 ratio: 1.73205

Fri Jun  9 13:36:59 PDT 2017

 u^2 - 3 v^2 = 4

jagy@phobeusjunior:~$ 
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